Sylow's Theorem Explanation

Question

Can someone explain it to me? I've been working out of Galian's Contemporary Abstract Algebra this semester, but came into possession a copy of Dummit and Foote's book, which I am aware is substantially more advanced. It was there I stumbled upon Sylow's Theorem, and although I followed the harder book well up to that point, I don't get the theorem at all.

Any help would be appreciated.

A statement of the theorem in question:

If $P$ is a Sylow $p$-subgroup of $G$ and $Q$ is any $p$-subgroup of $G,$ then there exists $g \in$ G such that $Q \leq gPg^{-1}.$ In particular, any two Sylow $p$-subgroups of $G$ are conjugate in $G$.

Answer 1

So it would definitely be helpful if you could clarify what part of this theorem doesn't make sense to you. As a broad overview it is saying if $\vert G \vert = p^m r$ and $p$ doesn't divide $r$ then if you have a subgroup $P$ with $\vert P \vert = p^m$ and any subgroup $Q$ where $\vert Q \vert = p^{m-k}$ then there is an element of the conjugacy class of $Q$ that is a subgroup of $P$. Using this and the pigeon hole principle it them follows that any Sylow $p$-subgroups are conjugates.

EDIT:

Seeing that you were having trouble with the proof and not the concept I went to my copy of Dummit and Foote to see what they do. The basic structure of their proof of this part of Sylow's Theorem is that they assume it is false and derive a contradiction. The crucial facts they use is that if $\mathcal{O}_i$ denotes an orbit of $P_i$, a conjugate of some Sylow $p$-subgroup $P$, under the action of some subgroup $Q$ of $G$ by conjugation then $$\vert \mathcal{O}_i \vert = \vert Q : P_i \cap Q \vert$$ and that the sum of the orders of all the orbits $r$ is $1 \pmod p$.

For the proof we assume that $Q$ is not contained in any $P_i$ (or isn't in a conjugate of $P$) and this imediately tells us that $\vert Q : P_i \cap Q \vert > 1$ for every $i$ (because if it wasn't then $Q$ would be contained in $P_i$. We know that (since these are finite groups) $$\vert Q : P_i \cap Q \vert = \frac{ \vert Q \vert}{\vert P_i \cap Q \vert }$$ and so (since $Q$ is a $p$ group and this fraction isn't $1$) $p$ divides the index of $P_i \cap Q$ in $Q$ and also the size of the orbit $\mathcal{O}_i$. Now since this is true for every $\mathcal{O}_i$ $p$ also divides the sum $$ r = \mathcal{O}_1 + \cdots + \mathcal{O}_n$$ however this contradict the fact that $r \equiv 1 \pmod p$. Therefore the assumption that $Q$ is not contained in a conjugate of $P$ must be false.

Answer 2

Let $Q$ act by left-multiplication on $\mathcal{S}$, the set of left cosets of $P$. Note that the cardinality of $\mathcal{S}$ is index$[G:P]$ and hence not divisible by $p$. The length of the orbit of $gP$ is index$[Q:Q \cap P^{g^{-1}}]$, a $p$-power. Since $|\mathcal{S}|$ is the sum of these orbit lengths and $p \nmid |\mathcal{S}|$, there must be an orbit of length $1$, which means $Q \subseteq P^{g^{-1}}$ for some $g \in G$.